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Chapter 9: Problem 38

Let \(A=\left[\begin{array}{rr}2 & -3 \\ 5 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]\). Find \(f(A), g(A), f(B), g(B)\), where \(f(t)=2 t^{2}-5 t+6\) and \(g(t)=t^{3}-2 t^{2}+t+3\)

Short Answer

Expert verified
In conclusion, the values of f(A), g(A), f(B), and g(B) are: \[ f(A) = \begin{bmatrix} 4 & 27 \\ 26 & 3 \end{bmatrix}, \quad g(A) = \begin{bmatrix} 1 & -8 \\ 73 & 1 \end{bmatrix}, \quad f(B) = \begin{bmatrix} 3 & 4 \\ 6 & 9 \end{bmatrix}, \quad g(B) = \begin{bmatrix} 1 & 1 \\ 3 & 9 \end{bmatrix} \]

Step by step solution

01

Evaluate f(t) and g(t) for each element of Matrix A

First, we will evaluate f(t) and g(t) for each element of matrix A as: \( f(2) = 2(2)^2 - 5(2) + 6 = 4\) \( f(-3) = 2(-3)^2 - 5(-3) + 6 = 27\) \( f(5) = 2(5)^2 - 5(5) + 6 = 26\) \( f(1) = 2(1)^2 - 5(1) + 6 = 3\) \(g(2) = 2^3 - 2(2)^2 + 2 + 3 = 1\) \(g(-3) = (-3)^3 - 2(-3)^2 + (-3) + 3 = -8\) \(g(5) = 5^3 - 2(5)^2 + 5 + 3 = 73\) \(g(1) = 1^3 - 2(1)^2 + 1 + 3 = 1\)
02

Write the resultant matrices f(A) and g(A)

Now that we have evaluated f(t) and g(t) for each element of matrix A, let's write the resultant matrices f(A) and g(A): Matrix f(A) is: \[ \begin{bmatrix} 4 & 27 \\ 26 & 3 \end{bmatrix} \] Matrix g(A) is: \[ \begin{bmatrix} 1 & -8 \\ 73 & 1 \end{bmatrix} \]
03

Evaluate f(t) and g(t) for each element of Matrix B

Next, we will evaluate f(t) and g(t) for each element of matrix B as: \( f(1) = 2(1)^2 - 5(1) + 6 = 3\) \( f(2) = 2(2)^2 - 5(2) + 6 = 4\) \( f(0) = 2(0)^2 - 5(0) + 6 = 6\) \( f(3) = 2(3)^2 - 5(3) + 6 = 9\) \(g(1) = 1^3 - 2(1)^2 + 1 + 3 = 1\) \(g(2) = 2^3 - 2(2)^2 + 2 + 3 = 1\) \(g(0) = 0^3 - 2(0)^2 + 0 + 3 = 3\) \(g(3) = 3^3 - 2(3)^2 + 3 + 3 = 9\)
04

Write the resultant matrices f(B) and g(B)

Now that we have evaluated f(t) and g(t) for each element of matrix B, let's write the resultant matrices as f(B) and g(B): Matrix f(B) is: \[ \begin{bmatrix} 3 & 4 \\ 6 & 9 \end{bmatrix} \] Matrix g(B) is: \[ \begin{bmatrix} 1 & 1 \\ 3 & 9 \end{bmatrix} \] In conclusion, the values of f(A), g(A), f(B), and g(B) are respectively: \[ f(A) = \begin{bmatrix} 4 & 27 \\ 26 & 3 \end{bmatrix}, \quad g(A) = \begin{bmatrix} 1 & -8 \\ 73 & 1 \end{bmatrix}, \quad f(B) = \begin{bmatrix} 3 & 4 \\ 6 & 9 \end{bmatrix}, \quad g(B) = \begin{bmatrix} 1 & 1 \\ 3 & 9 \end{bmatrix} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Evaluation
Matrix evaluation is the process of applying a given function to each element within the matrix. In our example, we have two matrices, \( A \) and \( B \), and two polynomial functions, \( f(t) \) and \( g(t) \). Matrix evaluation involves substituting each element of the matrices into the polynomials and calculating the result. For example, if you have a polynomial \( f(t) = 2t^2 - 5t + 6 \), and you want to evaluate it for a matrix element like 2, you substitute to get \( f(2) = 2(2)^2 - 5(2) + 6 \), which equals 4. This substitution and evaluation are performed for each element in the matrices A and B. This method allows us to transform matrices by applying operations that may represent physical processes or data transformations.
Polynomial Functions
Polynomial functions are expressions consisting of variables and coefficients involving only addition, subtraction, multiplication, and non-negative integer exponents of variables. In this exercise, the polynomial functions \( f(t) = 2t^2 - 5t + 6 \) and \( g(t) = t^3 - 2t^2 + t + 3 \) are applied to each element of matrices \( A \) and \( B \), respectively.

Polynomial functions are fundamental in algebra because they exhibit continuous and smooth curves when graphed. Knowing how to evaluate polynomial functions at specific points, like the elements of a matrix, enables us to model and solve various mathematical problems. Evaluating polynomials is done systematically by replacing every occurrence of the variable with the number or element you wish to substitute, then performing arithmetic operations following the order of operations.
Matrix Algebra
Matrix algebra involves a set of mathematical rules and operations that apply to matrices, which can represent systems of equations, transformations, and more. In matrix algebra, you can perform standard arithmetic operations such as addition, subtraction, and multiplication, as well as more complex operations like finding determinants or eigenvalues.

In the exercise given, matrix algebra is employed to evaluate functions at each element in the matrices \( A \) and \( B \), and to express the results in new matrices. The process demonstrates how matrix operations can be extended beyond basic arithmetic into function evaluation, showing how complex matrices can be manipulated to yield new solutions or interpretations in applied mathematics. Understanding matrix algebra is crucial in fields like engineering, physics, and computer science where they are used extensively to model real-world systems.

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Most popular questions from this chapter

Prove Theorem 9.1: Let \(f\) and \(g\) be polynomials. For any square matrix \(A\) and scalar \(k\), (i) \((f+g)(A)=f(A)+g(A)\), (iii) \((k f)(A)=k f(A)\), (ii) \((f g)(A)=f(A) g(A)\), (iv) \(f(A) g(A)=g(A) f(A)\) Suppose \(f=a_{n} t^{n}+\cdots+a_{1} t+a_{0}\) and \(g=b_{m} t^{m}+\cdots+b_{1} t+b_{0} .\) Then, by definition, $$ f(A)=a_{n} A^{n}+\cdots+a_{1} A+a_{0} I \quad \text { and } \quad g(A)=b_{m} A^{m 1}+\cdots+b_{1} A+b_{0} I $$ (i) Suppose \(m \leq n\) and let \(b_{i}=0\) if \(i>m\). Then $$ f+g=\left(a_{n}+b_{n}\right) t^{n}+\cdots+\left(a_{1}+b_{1}\right) t+\left(a_{0}+b_{0}\right) $$ Hence, $$ \begin{aligned} (f+g)(A) &=\left(a_{n}+b_{n}\right) A^{n}+\cdots+\left(a_{1}+b_{1}\right) A+\left(a_{0}+b_{0}\right) I \\ &=a_{n} A^{n}+b_{n} A^{n}+\cdots+a_{1} A+b_{1} A+a_{0} I+b_{0} I=f(A)+g(A) \end{aligned} $$ (ii) By definition, \(f g=c_{n+m} t^{n+m}+\cdots+c_{1} t+c_{0}=\sum_{k=0}^{n+m} c_{k} t^{k}\), where $$ c_{k}=a_{0} b_{k}+a_{1} b_{k-1}+\cdots+a_{k} b_{0}=\sum_{i=0}^{k} a_{i} b_{k-i} $$ Hence, \((f g)(A)=\sum_{k=0}^{n+m} c_{k} A^{k}\) and $$ f(A) g(A)=\left(\sum_{i=0}^{n} a_{i} A^{i}\right)\left(\sum_{j=0}^{m} b_{j} A^{j}\right)=\sum_{i=0}^{n} \sum_{j=0}^{m} a_{i} b_{j} A^{i+j}=\sum_{k=0}^{n+m} c_{k} A^{k}=(f g)(A) $$ (iii) By definition, \(k f=k a_{n} t^{n}+\cdots+k a_{1} t+k a_{0}\), and so $$ (k f)(A)=k a_{n} A^{n}+\cdots+k a_{1} A+k a_{0} I=k\left(a_{n} A^{n}+\cdots+a_{1} A+a_{0} I\right)=k f(A) $$ (iv) By (ii), \(g(A) f(A)=(g f)(A)=\left(f_{B}\right)(A)=f(A) g(A)\)

Let \(q(x, y)=x^{2}+6 x y-7 y^{2} .\) Find an orthogonal substitution that diagonalizes \(q\). Find the symmetric matrix \(A\) that represents \(q\) and its characteristic polynomial \(\Delta(t) .\) We have $$ A=\left[\begin{array}{rr} 1 & 3 \\ 3 & -7 \end{array}\right] \quad \text { and } \quad \Delta(t)=t^{2}+6 t-16=(t-2)(t+8) $$ The eigenvalues of \(A\) are \(\lambda=2\) and \(\lambda=-8\). Thus, using \(s\) and \(t\) as new variables, a diagonal form of \(q\) is $$ q(s, t)=2 s^{2}-8 t^{2} $$ The corresponding orthogonal substitution is obtained by finding an orthogonal set of eigenvectors of \(A\). (i) Subtract \(\lambda=2\) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{rr} -1 & 3 \\ 3 & -9 \end{array}\right], \quad \text { corresponding to } \begin{array}{r} -x+3 y=0 \\ 3 x-9 y=0 \end{array} \text { or }-x+3 y=0 $$ A nonzero solution is \(u_{1}=(3,1)\) (ii) Subtract \(\lambda=-8\) (or add 8) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{ll} 9 & 3 \\ 3 & 1 \end{array}\right], \quad \text { corresponding to } \quad \begin{aligned} &9 x+3 y=0 \\ &3 x+y=0 \end{aligned} \text { or } 3 x+y=0 $$ A nonzero solution is \(u_{2}=(-1,3)\). As expected, because \(A\) is symmetric, the eigenvectors \(u_{1}\) and \(u_{2}\) are orthogonal. Now normalize \(u_{1}\) and \(u_{2}\) to obtain, respectively, the unit vectors $$ \hat{u}_{1}=(3 / \sqrt{10}, 1 / \sqrt{10}) \quad \text { and } \quad \hat{u}_{2}=(-1 / \sqrt{10}, 3 / \sqrt{10}) $$ Finally, let \(P\) be the matrix whose columns are the unit vectors \(\hat{u}_{1}\) and \(\hat{u}_{2}\), respectively, and then \([x, y]^{T}=P[s, t]^{T}\) is the required orthogonal change of coordinates. That is, $$ \left.P=\mid \begin{array}{cc} 3 / \sqrt{10} & -1 / \sqrt{10} \\ 1 / \sqrt{10} & 3 / \sqrt{10} \end{array}\right] \quad \text { and } \quad x=\frac{3 s-t}{\sqrt{10}}, \quad y=\frac{s+3 t}{\sqrt{10}} $$ One can also express \(s\) and \(t\) in terms of \(x\) and \(y\) by using \(P^{-1}=P^{T} .\) That is, $$ s=\frac{3 x+y}{\sqrt{10}}, \quad t=\frac{-x+3 t}{\sqrt{10}} $$

Let \(A\) be any square matrix and let \(f(t)\) be any polynomial. Prove (a) \(\left(P^{-1} A P\right)^{n}=P^{-1} A^{n} P\). (b) \(f\left(P^{-1} A P\right)=P^{-1} f(A) P\) (c) \(f\left(A^{T}\right)=[f(A)]^{T}\) (d) If \(A\) is symmetric, then \(f(A)\) is symmetric.

Let \(A=\left[\begin{array}{rr}7 & 3 \\ 3 & -1\end{array}\right]\). Find an orthogonal matrix \(P\) such that \(D=P^{-1} A P\) is diagonal.First find the characteristic polynomial \(\Delta(t)\) of \(A\). We have $$ \Delta(t)=t^{2}-\operatorname{tr}(A) t+|A|=t^{2}-6 t-16=(t-8)(t+2) $$ Thus, the eigenvalues of \(A\) are \(\lambda=8\) and \(\lambda=-2\). We next find corresponding eigenvectors. Subtract \(\lambda=8\) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{rr} -1 & 3 \\ 3 & -9 \end{array}\right], \quad \text { corresponding to } \quad \begin{array}{r} -x+3 y=0 \\ 3 x-9 y=0 \end{array} \text { or } x-3 y=0 $$ A nonzero solution is \(u_{1}=(3,1)\) Subtract \(\lambda=-2\) (or add 2 ) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{ll} 9 & 3 \\ 3 & 1 \end{array}\right] \text {, corresponding to } \quad \begin{aligned} &9 x+3 y=0 \\ &3 x+y=0 \end{aligned} \quad \text { or } \quad 3 x+y=0 $$ A nonzero solution is \(u_{2}=(1,-3)\). As expected, because \(A\) is symmetric, the eigenvectors \(u_{1}\) and \(u_{2}\) are orthogonal. Normalize \(u_{1}\) and \(u_{2}\) to obtain, respectively, the unit vectors $$ \hat{u}_{1}=(3 / \sqrt{10}, 1 / \sqrt{10}) \quad \text { and } \quad \hat{u}_{2}=(1 / \sqrt{10},-3 / \sqrt{10}) $$ Finally, let \(P\) be the matrix whose columns are the unit vectors \(\hat{u}_{1}\) and \(\hat{u}_{2}\), respectively. Then $$ P=\left[\begin{array}{cc} 3 / \sqrt{10} & 1 / \sqrt{10} \\ 1 / \sqrt{10} & -3 / \sqrt{10} \end{array}\right] \quad \text { and } \quad D=P^{-1} A P=\left[\begin{array}{rr} 8 & 0 \\ 0 & -2 \end{array}\right] $$ As expected, the diagonal entries in \(D\) are the eigenvalues of \(A\).

Let \(B=\left[\begin{array}{rrr}8 & 12 & 0 \\ 0 & 8 & 12 \\ 0 & 0 & 8\end{array}\right]\). Find a real matrix \(A\) such that \(B=A^{3}\).
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